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题目链接:http://acm.uestc.edu.cn/#/problem/show/1226
题目大意就是构造一个行列和每个角的2*2都是1234的4*4矩阵。
用dfs暴力搜索,不过需要每一步进行判断是否已经出现了重复,如果最后再判断的话复杂度有点高。
代码:
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <algorithm> #include <set> #include <map> #include <queue> #include <string> #define LL long long using namespace std; char str[5][5]; int a[5][5]; bool vis[5]; bool flag; bool judge(int x[][5]) { for (int i = 0; i < 4; ++i) { memset(vis, false, sizeof(vis)); for (int j = 0; j < 4; ++j) { if (!x[i][j]) continue; if (vis[x[i][j]]) return false; else vis[x[i][j]] = true; } } for (int j = 0; j < 4; ++j) { memset(vis, false, sizeof(vis)); for (int i = 0; i < 4; ++i) { if (!x[i][j]) continue; if (vis[x[i][j]]) return false; else vis[x[i][j]] = true; } } for (int i = 0; i <= 2; i += 2) { for (int j = 0; j <= 2; j += 2) { memset(vis, false, sizeof(vis)); for (int xx = 0; xx <= 1; xx++) { for (int yy = 0; yy <= 1; yy++) { if (!x[i+xx][j+yy]) continue; if (vis[x[i+xx][j+yy]]) return false; else vis[x[i+xx][j+yy]] = true; } } } } return true; } void input() { for (int i = 0; i < 4; ++i) { scanf("%s", str[i]); for (int j = 0; j < 4; ++j) { if (str[i][j] == ‘*‘) a[i][j] = 0; else a[i][j] = str[i][j]-‘0‘; } } } void dfs(int i, int j) { if (flag) return; i += j/4; j %= 4; if (i == 4) { if (judge(a)) { for (int i = 0; i < 4; ++i) { for (int j = 0; j < 4; ++j) printf("%d", a[i][j]); printf("\n"); } flag = true; } return; } if (!judge(a)) return; if (!a[i][j]) { for (int x = 1; x <= 4; ++x) { a[i][j] = x; dfs(i, j+1); a[i][j] = 0; } } else dfs(i, j+1); } void work() { flag = false; dfs(0, 0); } int main() { //freopen("test.in", "r", stdin); int T; scanf("%d", &T); for (int times = 1; times <= T; ++times) { printf("Case #%d:\n", times); input(); work(); } return 0; }
ACM学习历程—UESTC 1222 Sudoku(矩阵)(2015CCPC H)
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原文地址:http://www.cnblogs.com/andyqsmart/p/4997382.html